prove that $18^n-n^{16}$ is divisible by $17$ for any natural $n$

Question

prove that

$18^n-n^{16}$ is divisible by $17$ for any natural $n$.

I have tried using Fermat's small theorem so far and got nowhere as I get contradiction for the case $n=17$ either way and for any case $n\neq16k$ I have no clue what to do with it.

Well, you can't do it if it's false. It's true when $n$ is not divisible by $17$ though. — Mark, Oct 18 '23 at 18:34
Any reason why you have typed $16$ in a weird, hard to read way? — FShrike, Oct 18 '23 at 18:37
yes, i realize it works for n≠17k, but how do i prove this — Delta Account, Oct 18 '23 at 18:39
@DeltaAccount You already wrote that you should use Fermat's little theorem, so just try it. Check what $18^n$ and $n^{16}$ are equal to mod $n$. — Mark, Oct 18 '23 at 18:40
because I couldn't use $s to indicate the power of two digit nums — Delta Account, Oct 18 '23 at 18:40
Isn't it "Fermat's Little Theorem", as a play on "Fermat's Last Theorem"? — Shaun, Oct 18 '23 at 18:57
The proof when $,17\nmid n,$ is an immediate consequence of the linked congruence laws, i.e. $18\equiv 1\Rightarrow 18^n\equiv 1^n\equiv 1,,$ and $,n\not\equiv 0\Rightarrow n^{16}\equiv 1,$ by Fermat. — Bill Dubuque, Oct 18 '23 at 19:11

score -1 · Accepted Answer · answered Oct 18 '23 at 18:57

-1

As someone mentioned, this is true when n is not divisible by 17! So in that case, we need to look at this whole equation mod 17, and we will prove it is equal to 0. First, $$18^n - n^{16} \mod{17} = (18 \mod{17})^n - (n \mod{17})^{16} \\= 1^n-n^{16} \mod17 = 1 - n^{16} \mod17 $$ so no we just need to show that for any $n$ which is not divisible by 17 ( $n \neq 0) \mod17)$), the following equality holds: $$ n^{16} = 1 \mod17$$ And because 17 is prime, such $n$ is invertible, and we can use the Fermat little theorem which says that $n^17 = n \mod17$ and get that $$n^{16} = n^{17} * n_{-1} = n * n^{-1} = 1 \mod17$$

answered Oct 18 '23 at 18:57

Nadav Kalma

181
5

Please strive not to post more (dupe) answers to dupes of FAQs, cf. recent site policy announcement here. Please delete the answer so that the dupe can be more easily cleaned up. – Bill Dubuque Oct 18 '23 at 19:15
@BillDubuque But how is this question a duplicate of that other question in the link? Not only it's not the same problem, the other question doesn't even deal with division by $17$. Not sure how reading the answers there will help OP solve this problem. (I hope OP already solved it anyway) – Mark Oct 18 '23 at 20:13
@Mark See my closure comment and see abstract duplicates. – Bill Dubuque Oct 18 '23 at 20:28

prove that $18^n-n^{16}$ is divisible by $17$ for any natural $n$

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